Two dielectric slabs of dielectric constants `K_(1) and K_(2)` are filled in between the two plates, eachof area A of the parallel plate capacitor as shown in the figure. Find the net capacitance of the capacitor.

Explanation

The arrangement is equivalent to a parallel of two capacitors, each with plate area `A/2` and separation d. So net capacitance is
`C=C_(1)+C_(2) =(epsi_(0) ""A/2 K_(1))/(d) (epsi_(0)""A/2 K_(2))/(d)`
`=(epsi_(0) A(K_(1)+K_(2)))/(2d)`